= x^2 - 7xy + 10y^2
= x^2 - 2xy -5xy + 10y^2
= x( x - 2y ) - 5y( x - 2y )
= ( x - 2y )( x - 5y )

x² – 7xy + 10y2 

= x² – 2xy – 5xy + 10y²

= x(x – 2y) – 5y(x – 2y)

= (x – 2y)(x – 5y)

a)  \(x^2+6x+8\)

\(=\left(x^2-2x\right)-4x+8\)

\(=x\left(x-2\right)-4\left(x-2\right)\)

\(\left(x-2\right)\left(x-4\right)\)

b) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-5y\right)\)

a) x² - 6x + 8

= x² -2x - 4x +8

= x( x-2) -4( x-2)

= ( x-2)(x-4)

a/ \(x^2-6x+8=x^2-2x-4x+8\) \(8\) \(=x\left(x-2\right)-4\left(x-2\right)\) \(=\left(x-4\right)\left(x-2\right)\)

b/ \(x^2-7xy+10y^2\) \(=x^2-2xy-5xy+10y^2\) \(=x\left(x-2y\right)-5y\left(x-2y\right)\) \(=\left(x-5y\right)\left(x-2y\right)\)

$=x^2-2xy-5xy+10y^2\\=x(x-2y)-5y(x-2y)\\=(x-5y)(x-2y)$

x^2 - 7xy + 10y^2 
= (x^2 - 2xy) - (5xy - 10y^2) 
= x(x - 2y) - 5y( x - 2y) 
= (x - 5y)(x - 2y) 

mik mới học lớp 7 thôi!

Ta có : x³ - 3x² + 1 - 3x

= x³ - 2x² - x - (x² - 2x - 1) 

= x(x² - 2x - 1) - (x² - 2x - 1) 

= (x² - 2x - 1) (x - 1) 

a) x³ ​- 4x² - 12x + 27

= \(\left(x^3+3x^2\right)-\left(7x^2+21x\right)+\left(9x+27\right)\)

= \(\left(x+3\right)\left(x^2-7x+9\right)\)

b) 9x² + 6x - 8 

=\(9x^2-6x+12x-8=3x\left(3x-2\right)+4\left(3x-2\right)\)

=\(\left(3x-2\right)\left(3x+4\right)\)

c) x² - 7xy + 10y²

^{=\(x^2-5xy-2xy+10y^2=x\left(x-5y\right)-2y\left(x-5y\right)\)}

^{=\(\left(x-5y\right)\left(x-2y\right)\)}

a) x³ ​- 4x² - 12x + 27

 =x³ + 3x² - 7x² - 21x + 9x + 27 

= x²(x+3) - 7x(x+3) + 9(x+3)  

= (x² - 7x + 9)(x + 3)

b) 9x² + 6x - 8 

= 9x² - 6x + 12x - 8

= 3x(3x - 2) +  4(3x - 2) 

= (3x + 4)(3x - 2)

c) x² - 7xy + 10y²

= x² - 5xy - 2xy + 10y²

= x(x - 5y) - 2y(x - 5y)

= (x - 2y)(x - 5y)

d)  x⁸ + x⁷ + 1 

Ta thêm vào các số hạng x⁶, x⁵, x⁴, x³, x², x và cùng bớt đi các số hạng ấy ta có:

= x⁸ - x⁶ + x⁵ - x³ + x² + x⁷ - x⁵ + x⁴ -x² +x + x⁶ - x⁴ + x³ - x + 1

= x²(x⁶ - x⁴ + x³ - x + 1) + x(x⁶ - x⁴ + x³ - x + 1) + x⁶ - x⁴ + x³ - x + 1

= (x² + x + 1)(x⁶ - x⁴ + x³ - x + 1)

a) \(x^3-4x^2-12x+27\)

\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x+9-4x\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

b) \(9x^2+6x-8\)

\(=9x^2+6x+1-9\)

\(=\left(3x+1\right)^2-9\)

\(=\left(3x+1-9\right)\left(3x+1+9\right)\)

\(=\left(3x+8\right)\left(3x+10\right)\)

c) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-5\right)\)

a: x^2+4xy-21y^2

\(=x^2+7xy-3xy-21y^2\)

\(=x\left(x+7y\right)-3y\left(x+7y\right)\)

\(=\left(x+7y\right)\left(x-3y\right)\)

b: \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

=5x(x+y)+y(x+y)

=(x+y)(5x+y)

c: \(x^2+2xy-15y^2\)

\(=x^2+5xy-3xy-15y^2\)

=x(x+5y)-3y(x+5y)

=(x+5y)(x-3y)

d: \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

=x(x-2y)-5y(x-2y)

=(x-2y)(x-5y)

a) \(x^2+4xy-21y^2\)

\(=x^2+7xy-3xy-21y^2\)

\(=x\left(x+7y\right)-3y\left(x+7y\right)\)

\(=\left(x+7y\right)\left(x-3y\right)\)

b) \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

\(=5x\left(x+y\right)+y\left(x+y\right)\)

\(=\left(5x+y\right)\left(x+y\right)\)

c) \(x^2+2xy-15y^2\)

\(=x^2+5xy-3xy-15y^2\)

\(=x\left(x+5y\right)-3y\left(x+5y\right)\)

\(=\left(x+5y\right)\left(x-3y\right)\)

d) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-5y\right)\left(x-2y\right)\)

a) x² + 4xy - 21y²

= x² - 3xy + 7xy - 21y²

= (x² - 3xy) + (7xy - 21y²)

= x(x - 3y) + 7y(x - 3y)

= (x - 3y)(x + 7y)

b) 5x² + 6xy + y²

= 5x² + 5xy + xy + y²

= (5x² + 5xy) + (xy + y²)

= 5x(x + y) + y(x + y)

= (x + y)(5x + y)

c) x² + 2xy - 15y²

= x² + 2xy + y² - 16y²

= (x² + 2xy + y²) - 16y²

= (x + y)² - (4y)²

= (x + y - 4y)(x + y + 4y)

= (x - 3y)(x + 5y)

d) x² - 7xy + 10y²

= x² - 2xy - 5xy + 10y²

= (x² - 2xy) - (5xy + 10y²)

= x(x - 2y) - 5y(x - 2y)

= (x - 2y)(x - 5y)

\(x^2+4x-y^2+4\)

\(=\left(x^2+4x+4\right)-y^2\)

\(=\left(x^2+2\right)^2-y^4\)

\(=\left(x^2+y^2+2\right)\left(x^2-y^2+2\right)\)

\(\left(x^2+4x+4\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left(x+2+y\right)\left(x+2-y\right)\)

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hk tôt